HDU 5389 Zero Escape
来源:互联网 发布:客户数据库 编辑:程序博客网 时间:2024/06/05 11:06
Zero Escape
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Problem Description
Zero Escape, is a visual novel adventure video game directed by Kotaro Uchikoshi (you may hear about ever17?) and developed by Chunsoft.
Stilwell is enjoying the first chapter of this series, and in this chapter digital root is an important factor.
This is the definition of digital root on Wikipedia:
The digital root of a non-negative integer is the single digit value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached.
For example, the digital root of65536 is 7 , because 6+5+5+3+6=25 and 2+5=7 .
In the game, every player has a special identifier. Maybe two players have the same identifier, but they are different players. If a group of players want to get into a door numberedX(1≤X≤9) , the digital root of their identifier sum must be X .
For example, players{1,2,6} can get into the door 9 , but players {2,3,3} can't.
There is two doors, numberedA and B . Maybe A=B , but they are two different door.
And there isn players, everyone must get into one of these two doors. Some players will get into the door A , and others will get into the door B .
For example:
players are{1,2,6} , A=9 , B=1
There is only one way to distribute the players: all players get into the door9 . Because there is no player to get into the door 1 , the digital root limit of this door will be ignored.
Given the identifier of every player, please calculate how many kinds of methods are there,mod 258280327 .
Stilwell is enjoying the first chapter of this series, and in this chapter digital root is an important factor.
This is the definition of digital root on Wikipedia:
The digital root of a non-negative integer is the single digit value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached.
For example, the digital root of
In the game, every player has a special identifier. Maybe two players have the same identifier, but they are different players. If a group of players want to get into a door numbered
For example, players
There is two doors, numbered
And there is
For example:
players are
There is only one way to distribute the players: all players get into the door
Given the identifier of every player, please calculate how many kinds of methods are there,
Input
The first line of the input contains a single number T , the number of test cases.
For each test case, the first line contains three integersn , A and B .
Next line containsn integers idi , describing the identifier of every player.
T≤100 , n≤105 , ∑n≤106 , 1≤A,B,idi≤9
For each test case, the first line contains three integers
Next line contains
Output
For each test case, output a single integer in a single line, the number of ways that these n players can get into these two doors.
Sample Input
43 9 11 2 63 9 12 3 35 2 31 1 1 1 19 9 91 2 3 4 5 6 7 8 9
Sample Output
101060
思路:
用动态规划,因为各位数总和是能是1~9,这里用dp[i]][j]表示前i个数总和为j的所有可能数(这个和是个位数)。
注意:
(X+Y)% 9 = (X%9 + Y%9) % 9;当值为0时置为9。
转移方程:
dp[i][j] = dp[i-1][j] + dp[i-1][j-id[i]] (j-id[i] >= 0)
dp[i][j] = dp[i-1][j] + dp[i-1][j-id[i]+9] (j-id[i] < 0)
注意最终如果A+B != SUM,则答案为0,否则只要考虑dp[n][A],因为剩下的都得放到B里面,还要判断一下A、B和SUM的关系。
0 0
- hdu 5389 Zero Escape
- HDU 5389 Zero Escape
- HDU 5389 Zero Escape
- hdu 5389 Zero Escape
- HDU 5389 Zero Escape
- HDU 5389 Zero Escape
- hdu 5389 Zero Escape(dp)
- hdu 5389 Zero Escape dp
- hdu 5389 Zero Escape (dp)
- hdu 5389 Zero Escape (dp)
- hdu 5389 Zero Escape(dp)
- HDU 5389 Zero Escape(DP + 滚动数组)
- hdu 5389 Zero Escape(dp)
- hdu 5389 Zero Escape(dp类似背包)
- 计数DP(Zero Escape,HDU 5389)
- 【HDOJ 5389】 Zero Escape
- HDU 5389 Zero Escape(dp解法详解) 已更新
- HDU 5389 Zero Escape(dp啊 多校)
- ADV-92-算法提高-求最大公约数.
- Java泛型的上限和下限
- 单一入口文件的定义
- B
- XMPP协议的原理介绍
- HDU 5389 Zero Escape
- 3n+1问题
- JS
- [java]java实现哈夫曼编码
- 欢迎使用CSDN-markdown编辑器
- 使用showOptionDialog显示多项选择框
- 【死磕Java并发】—–J.U.C之Condition
- 棋子翻转
- 数据库表操作